Question

Daniel Reisman of Niverville, NY submitted the followine question to Marilyn vos Savant's December 27, 1998, Parade Magazine column, "Ask Marilyn." At a monthly 'casino night,' there is a game called Chuck-a-Luck. Three dice are rolled in a wire cage. You place a bet on any number from 1 to 6. If any one of the three dice comes up with your number, you win the amount of your bet. (You also get your onen stake back). If more than one die comes up with your number, you win the amount of your bet for each match. For example, if you had a $1 bet on number 5, and each of the dice came up with 5, you would win $3. It appears that the odds of winning are 1 in b for each of the three dice for a total of 3 out of 6 or 50%. Adding the possibility of having more than one die come up with your number, the odds would seem to be in the gambler's favor. What are the odds of winning this game? I can't believe that a casino game would favor the gambler. Daniel computed the probabilities incorrectly. There are four possible outcomes. (The selected number can match 0, 1, 2, or 3 of the dice.) The random variable X represents the profit from a $1 bet in Chuck-a-Luck. The table below summarizes the probabilities of earning a profit of x dollars from a $1 bet. Use this table to answer the following questions. Number of dice matching the chosen number O dice, 1 die, 2 dice, 3 dice. Profit x $-1, $1, $2, $3. Probability P(X=x) 125/216 75/216 15/216 1/216

g. What is the probability that a player who plays once will lose the game? In other words, what is the probability that a player will match none of the three dice?

h. What is the probability that a player will match all three of the dice?

i. What is the probability that a player will match at least one of the dice?

Answer 1

Total number of possibilities in sample space = 6×6×6 = 216

Probability distribution table is as follows :

Part a)

odds of winning the game by 1 matching = (1/6)×(5/6)×(5/6)×3C2 = 75/216

Odds of winning the game by 2 matching = (1/6)×(1/6)×(5/6) × 3C2 = 15/216

Odds of winning game by 3 matching = (1/6)×(1/6)×(1/6) = 1/216

So, total odds of winning = (75+15+1)/216 = 91/216.

g) Probability of losing the game = probability that none of the dice matches = (5/6)×(5/6)×(5/6) = 125/216

h) probability of matching all three dices = (1/6)×(1/6)×(1/6) =1/216

i) probability of matching at least one of the dice = 1- probability of matching zero dice

= 1- (125/216) = 91/216

Comment : this game is not favouring the gambler as odds of losing the game(125/216) is more than odds of winning the game(91/216).

Expected profit of this game = sum of (probabilities of winning × corresponding profit )

= (125/216)×(-1) + (75/216)×(1) + (15/216) ×(2) +(1/216)×(3)

= - 0.0787 dollar(negative)

Negative expected profit shows that the gambler or player is expected to lose 0.0787 dollar on each bet.

Hence the game is not favouring the gambler as all.